Answer:
5.004kg
Explanation:
Combustion of carbon
C+O2=CO2
from the relationship of molar ratio
mass of carbon/molar mass of carbon=volume of CO2 produced\molar vol(22.4 dm3)
mass of carbon =1000kg
atomic mass of carbon =12
volume of CO2 produced=1000×22.4/12
volume of CO2 produced =1866.6dm3
from the combustion reaction equation provided
CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)
applying the same relationship of molar ratio
no of mole of CO2=no of mole of urea
therefore
vol of CO2\22.4=mass of urea/molar mass of urea
molar mass of urea=60.06g/mol
from the first calculation
vol of CO2=1866.6dm3
mass of urea=1866.6×60.06/22.4
mass of urea=5004.82kg
Hey there!
C₉H₂O + O₂ → CO₂ + H₂O
First let's balance the C.
There's 9 on the left and 1 on the right. So, let's add a coefficient of 9 in front of CO₂.
C₉H₂O + O₂ → 9CO₂ + H₂O
Next let's balance the H.
There's 2 on the left and 2 on the right. This means it's already balanced.
C₉H₂O + O₂ → 9CO₂ + H₂O
Lastly, let's balance the O.
There's 3 on the left and 19 on the right. So, let's add a coefficient of 9 in front of O₂.
C₉H₂O + 9O₂ → 9CO₂ + H₂O
This is our final balanced equation.
Hope this helps!
We know that, M1V1 = M2V2
(Initial) (Final)
where, M1 and M2 are initial and final concentration of soution respectively.
V1 and V2 = initial and final volume of solution respectively
Given: M1 = 12 m, V1 = 35 ml and V2 = 1.2 l = 1200 ml
∴ M2 = M1V1/V2 = (12 × 35)/ 1200 = 0.35 m
Final concentration of solution is 0.35 m
